/*

 * Copyright 2000-2016 The OpenSSL Project Authors. All Rights Reserved.

 *

 * Licensed under the Apache License 2.0 (the "License").  You may not use

 * this file except in compliance with the License.  You can obtain a copy

 * in the file LICENSE in the source distribution or at

 * https://www.openssl.org/source/license.html

 */

#include "internal/cryptlib.h"

#include "bn_local.h"

/* least significant word */

#define BN_lsw(n) (((n)->top == 0) ? (BN_ULONG) 0 : (n)->d[0])

/* Returns -2 for errors because both -1 and 0 are valid results. */

int BN_kronecker(const BIGNUM *a, const BIGNUM *b, BN_CTX *ctx)

{

    int i;

    int ret = -2;               /* avoid 'uninitialized' warning */

    int err = 0;

    BIGNUM *A, *B, *tmp;

    /*-

     * In 'tab', only odd-indexed entries are relevant:

     * For any odd BIGNUM n,

     *     tab[BN_lsw(n) & 7]

     * is $(-1)^{(n^2-1)/8}$ (using TeX notation).

     * Note that the sign of n does not matter.

     */

    static const int tab[8] = { 0, 1, 0, -1, 0, -1, 0, 1 };

    bn_check_top(a);

    bn_check_top(b);

    BN_CTX_start(ctx);

    A = BN_CTX_get(ctx);

    B = BN_CTX_get(ctx);

    if (B == NULL)

        goto end;

    err = !BN_copy(A, a);

    if (err)

        goto end;

    err = !BN_copy(B, b);

    if (err)

        goto end;

    /*

     * Kronecker symbol, implemented according to Henri Cohen,

     * "A Course in Computational Algebraic Number Theory"

     * (algorithm 1.4.10).

     */

    /* Cohen's step 1: */

    if (BN_is_zero(B)) {

        ret = BN_abs_is_word(A, 1);

        goto end;

    }

    /* Cohen's step 2: */

    if (!BN_is_odd(A) && !BN_is_odd(B)) {

        ret = 0;

        goto end;

    }

    /* now  B  is non-zero */

    i = 0;

    while (!BN_is_bit_set(B, i))

        i++;

    err = !BN_rshift(B, B, i);

    if (err)

        goto end;

    if (i & 1) {

        /* i is odd */

        /* (thus  B  was even, thus  A  must be odd!)  */

        /* set 'ret' to $(-1)^{(A^2-1)/8}$ */

        ret = tab[BN_lsw(A) & 7];

    } else {

        /* i is even */

        ret = 1;

    }

    if (B->neg) {

        B->neg = 0;

        if (A->neg)

            ret = -ret;

    }

    /*

     * now B is positive and odd, so what remains to be done is to compute

     * the Jacobi symbol (A/B) and multiply it by 'ret'

     */

    while (1) {

        /* Cohen's step 3: */

        /*  B  is positive and odd */

        if (BN_is_zero(A)) {

            ret = BN_is_one(B) ? ret : 0;

            goto end;

        }

        /* now  A  is non-zero */

        i = 0;

        while (!BN_is_bit_set(A, i))

            i++;

        err = !BN_rshift(A, A, i);

        if (err)

            goto end;

        if (i & 1) {

            /* i is odd */

            /* multiply 'ret' by  $(-1)^{(B^2-1)/8}$ */

            ret = ret * tab[BN_lsw(B) & 7];

        }

        /* Cohen's step 4: */

        /* multiply 'ret' by  $(-1)^{(A-1)(B-1)/4}$ */

        if ((A->neg ? ~BN_lsw(A) : BN_lsw(A)) & BN_lsw(B) & 2)

            ret = -ret;

        /* (A, B) := (B mod |A|, |A|) */

        err = !BN_nnmod(B, B, A, ctx);

        if (err)

            goto end;

        tmp = A;

        A = B;

        B = tmp;

        tmp->neg = 0;

    }

 end:

    BN_CTX_end(ctx);

    if (err)

        return -2;

    else

        return ret;

}